Problem: Let $f$ be a differentiable function with $f(2)=-3$ and $f'(2)=-4$. What is the value of the approximation of $f(1.9)$ using the function's local linear approximation at $x=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-2.9$ (Choice B) B $-2.8$ (Choice C) C $-2.7$ (Choice D) D $-2.6$
Answer: The local linear approximation of $f$ at $x=2$ is achieved using the equation of the line tangent to $f$ at $x=2$. Let $L(x)$ represent this equation. We can find $L(x)$ using the general formula for the tangent to the graph of function $u$ at $x=a$ : $y=u'(a)(x-a)+u(a)$ [Is there a way to find this formula without memorizing?] In our case, $L(x)=f'(2)(x-2)+f(2)$. Plugging $f(2)=-3$ and $f'(2)=-4$, we obtain $L(x)=-4(x-2)-3$. To approximate $f(1.9)$, all we need is to plug $x=1.9$ into $L(x)$. $\begin{aligned} L(1.9)&=-4(1.9-2)-3 \\\\ &=-4(-0.1)-3 \\\\ &=-2.6 \end{aligned}$ In conclusion, the approximation of $f(1.9)$ using the function's local linear approximation at $x=2$ is $-2.6$.